A2.44 kg block is pushed 1.55 m up a vertical wall with constant speed by a constant force of magnitude f applied at an angle of 69.6 ◦ with the horizontal. the acceleration of gravity is 9.8 m/s 2 . if the coefficient of kinetic friction between the block and wall is 0.691, find the work done by f.

The work done is L= 49.83 Joules

Explanation:

m= 2.44kg

d= 1.55m

α= 69.6°

g= 9.8 m/s²

μ= 0.691

F= ?

Fy= F*sin(α)

Fx= F*cos(α)

Fr= μ * Fx

Fr= μ * F*cos(α)

W= m*g

W= 23.91 N

Fy - W - Fr = 0

F*sin(α) - W -  μ * F*cos(α) = 0

F* ( sin(α) - μ *cos(α) ) = W

F= W /  ( sin(α) - μ *cos(α) )

F= 34.3 N

L= Fy * d

L= 49.83 J

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